Let be a function such that . Then
- A.is one-one in but not in
- B.is one-one in
- C.is many-one in
- D.is many-one in
Show answer and solution
Correct option: A
Solution
f(x) = 1 + 2x/(x²+1). Differentiating: f'(x) = [2(x²+1) - 2x·2x]/(x²+1)² = 2(1-x²)/(x²+1)².
f'(x) > 0 for x ∈ (-1,1) and f'(x) < 0 for |x| > 1, so f increases then decreases — it is not one-one on all of ℝ (e.g. f(-∞) = 1 = f(∞), and f has the same value at symmetric points far from origin).
On [1,∞), f'(x) ≤ 0 (equals 0 only at x=1) and strictly decreasing thereafter, so f is strictly monotone (one-one) on [1,∞). Similarly on (-∞,-1) f is also strictly decreasing, so it is one-one there too — ruling out options 3 and 4. Option 1 is correct: f is one-one on [1,∞) but not on (-∞,∞).
f'(x) > 0 for x ∈ (-1,1) and f'(x) < 0 for |x| > 1, so f increases then decreases — it is not one-one on all of ℝ (e.g. f(-∞) = 1 = f(∞), and f has the same value at symmetric points far from origin).
On [1,∞), f'(x) ≤ 0 (equals 0 only at x=1) and strictly decreasing thereafter, so f is strictly monotone (one-one) on [1,∞). Similarly on (-∞,-1) f is also strictly decreasing, so it is one-one there too — ruling out options 3 and 4. Option 1 is correct: f is one-one on [1,∞) but not on (-∞,∞).
[JEE Main 2023 · 29 Jan · Shift 1]